2009/12/4 Michael Kay <mike(_at_)saxonica(_dot_)com>:
It's both verbose and inefficient compared with the much simpler
<xsl:variable name="xslmyn" select="document('')/*/namespace::myn" />
Thanks ! I'll fix this.
(..) But there's no way of disovering in an
included module where it was included from.
I read on w3c.org that "the resource located by the href attribute
value is parsed as an XML document, and the children of the
xsl:stylesheet element in this document replace the xsl:include
element in the including document."
Thus I thought the included code could access namespaces aliases of
the main xsl:stylesheet element. Anyway, I was clearly wrong.
However, it all seesm very long-winded to me. Why don't you just do
<xsl:template match="/">
<xsl:if test="not(myn:rootElement)">
<xsl:message terminate="yes">Root element must be named myn:rootElement
in the correct namespace</xsl:message>
I think I can not use this alternative, as my grammar is not the main
one : it defines additional attributes and elements that expand the
base grammars (to be clear : XSD and WADL are the base grammars).
Thus, I have no idea of what is the root element, which grammar it
fits, etc. In fact, I know it for *each* of my XSLs, of course, but
not in my case, as I was looking for a generic template to check the
namespace version.
Thanks a lot for your detailed answer and advice, Michael.
Regards,
Jan
Regards,
Michael Kay
http://www.saxonica.com/
http://twitter.com/michaelhkay
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